How do you find the derivative of $y = arcsin (3 x + 2)$ $y = arcsin(3x + 2)$ ?

Question

3608 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-23T14:57:45

The answer is $= \frac{3}{\sqrt{1 - {(3 x + 2)}^{2}}}$ $=3/sqrt(1-(3x+2)^2)$

We need ${cos}^{2} x + sin 2 x = 1$ $cos^2x+sin2x=1$

and $(sinx)'=cosx$

$y=arcsin(3x+2)$

$:. siny=3x+2$

Differentiating with respect to $x$

$(siny)'=(3x+2)'$

$cosydy/dx=3$

$dy/dx=3/cosy$

$cos^2y=1-sin^2y=1-(3x+2)^2$

$cosy=sqrt(1-(3x+2)^2)$

So,

$dy/dx=3/sqrt(1-(3x+2)^2)$

How do you find the derivative of y = arcsin(3x + 2)y=arcsin(3x+2)y = arcsin(3x + 2)?